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JEE • CHEMISTRY • P-BLOCK ELEMENTS

P-block elements practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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181 Medium
181 of 320
Analyzing +4 | -1
S-I: Solubility of noble gases in water decreases with increasing size of the noble gas.

S-II: Solubility of noble gases in water is due to dipole induced dipole interaction.
View answer and explanation
Correct option: D
S-I is false, S-II is true
Option D is correct because the solubility of noble gases in water increases with increasing size due to increased polarizability and dipole-induced dipole interactions are responsible for their solubility.
182 Medium
182 of 320
Analyzing +4 | -1
A white precipitate was formed when BaCl2\mathrm{BaCl}_2 was added to water extract of an inorganic salt. Further, a gas ' X\mathrm{X} ' with characteristic odour was released when the formed white precipitate was dissolved in dilute HCl\mathrm{HCl}. The anion present in the inorganic salt is
View answer and explanation
Correct option: B
SO32−\mathrm{SO}_3^{2-}
Option B (SO32−\mathrm{SO}_3^{2-}) is correct because BaCl2 precipitates sulfite ions as BaSO3, which releases SO2 gas with a characteristic odor upon reaction with HCl.
183 Medium
183 of 320
Understanding +4 | -1
Assertion (A) : Both H3PO3\mathrm{H}_3 \mathrm{PO}_3 and H3PO4\mathrm{H}_3 \mathrm{PO}_4 have the same number of hydrogen atoms but H3PO4\mathrm{H}_3 \mathrm{PO}_4 is a tribasic acid and H3PO3\mathrm{H}_3 \mathrm{PO}_3 is a dibasic acid.

Reason (R) : 1 mole of H3PO3\mathrm{H}_3 \mathrm{PO}_3 is neutralised by 2 moles of NaOH\mathrm{NaOH} while 1 mole of H3PO4\mathrm{H}_3 \mathrm{PO}_4 is neutralised by 3 moles of NaOH\mathrm{NaOH}.
View answer and explanation
Correct option: B
Both A\mathbf{A} and R\mathbf{R} are true but R\mathbf{R} is not the correct explanation of A\mathbf{A}.
Option B (Both A\mathbf{A} and R\mathbf{R} are true but R\mathbf{R} is not the correct explanation of A\mathbf{A}) is correct because both the assertion and the reason are true statements, but the reason doesn't directly explain why H3PO3\mathrm{H}_3\mathrm{PO}_3 is dibasic and H3PO4\mathrm{H}_3\mathrm{PO}_4 is tribasic.
184 Medium
184 of 320
Understanding +4 | -1
Which of the following is mixed anhydride?
View answer and explanation
Correct option: D
N2O4\mathrm{N}_2 \mathrm{O}_4
Option D (N2O4\mathrm{N}_2 \mathrm{O}_4) is correct because N2O4\mathrm{N}_2 \mathrm{O}_4 is a mixed anhydride as it gives a mixture of HNO2\mathrm{HNO}_2 and HNO3\mathrm{HNO}_3 upon hydrolysis.
185 Medium
185 of 320
Understanding +4 | -1
Which of the following is/are CORRECT from the given below?

(I) Brown ring test for nitrate depends on the ability of Fe2+{Fe}^{2+} to reduce nitrate to nitric oxide
(II) Red phosphorus is much less reactive than white phosphorus

(III) Red phosphorus readily catches fire in air to give white fumes

(IV) White phosphorus is a translucent white waxy solid
View answer and explanation
Correct option: D
I, II and IV
Option D (I, II and IV) is correct because statements I, II, and IV are correct regarding the brown ring test, reactivity of red phosphorus, and the nature of white phosphorus.
186 Medium
186 of 320
Understanding +4 | -1
NH3\mathrm{NH}_3 reacts with HCl\mathrm{HCl}. The correct reasons are

I. NH3\mathrm{NH}_3 can donate a pair of electrons.

II. The Cl−\mathrm{Cl}^{-} ion formed has a stable configuration.

III. The N\mathrm{N}-atom of NH3\mathrm{NH}_3 gains electrons.

IV. A proton in HCl\mathrm{HCl} can accept an electron pair from NH3\mathrm{NH}_3.

The correct option is
View answer and explanation
Correct option: B
I, II and IV
Option B (I, II and IV) is correct because ammonia donates an electron pair, chloride ion has a stable configuration, and a proton accepts the electron pair.
187 Medium
187 of 320
Understanding +4 | -1
A gas that cannot be collected over water is
View answer and explanation
Correct option: A
SO2S{O_2}
Option A (SO2S{O_2}) is correct because sulfur dioxide is highly soluble in water, reacting to form sulfurous acid, and therefore cannot be collected over water.
188 Medium
188 of 320
Remembering +4 | -1
When copper reacts with dilute nitric acid the gaseous product obtained is
View answer and explanation
Correct option: C
NO{NO}
Option C (NO{NO}) is correct because when copper reacts with dilute nitric acid, nitric oxide (NO{NO}) is the primary gaseous product.
189 Medium
189 of 320
Analyzing +4 | -1
What will be the A{A} and B{B} in the following equations.

8NH3+3Cl2(A)→6NH4Cl+N2NH3+3Cl2(B)→NCl3+3HCl\begin{gathered} 8{{N}}{{{H}}_3} + \mathop {3{{C}}{{{l}}_2}}\limits_{(A)} \to 6{{N}}{{\text{H}}_4}{{Cl}} + {{{N}}_2} \quad \\ {{N}}{{{H}}_3} + \mathop {3{{C}}{{{l}}_2}}\limits_{(B)} \to {{NC}}{{{l}}_3} + 3{{HCl}} \quad \\ \end{gathered}
View answer and explanation
Correct option: A
A={A}= Excess, B={B}= Excess
Option B (N2O5\mathrm{N}_2 \mathrm{O}_5) is correct because when N₂O₅ is hydrolyzed, it forms only HNO₃, where nitrogen remains in the +5 oxidation state, thus no disproportionation occurs.
190 Medium
190 of 320
Remembering +4 | -1
What is the correct order of occurrence (% by weight) in air of Ne, Ar and Kr?
View answer and explanation
Correct option: B
Ar>Ne>KrAr > Ne > Kr
Option B (Ar>Ne>KrAr > Ne > Kr) is correct because Argon has the highest percentage by weight in air, followed by Neon, and then Krypton.
191 Medium
191 of 320
Analyzing +4 | -1
Among the compounds given in the options, identify the one which does not gives a precipitate with excess of both NaOH\mathrm{NaOH} solution and NH3\mathrm{NH}_3 solution?
View answer and explanation
Correct option: B
Zn(NO3)2\mathrm{Zn}\left(\mathrm{NO}_3\right)_2
The correct answer is B (Zn(NO₃)₂) because zinc hydroxide initially precipitates with both NaOH and NH₃, but then dissolves in excess due to the formation of soluble complexes.
192 Medium
192 of 320
Analyzing +4 | -1
Which of the following is not correct?
View answer and explanation
Correct option: D
π∗2px<π2py<σ2pz\pi *2px < \pi 2py < \sigma 2pz (no. of nodal plane)
Option D (π∗2px<π2py<σ2pz\pi *2px < \pi 2py < \sigma 2pz (no. of nodal plane)) is correct because the number of nodal planes increases in the order σ2pz\sigma 2pz (0 nodal planes) < π2py\pi 2py (1 nodal plane) < π∗2px\pi *2px (2 nodal planes).
193 Medium
193 of 320
Understanding +4 | -1
Assertion (A): Phosphoric acid has no reducing properties.

Reason (R): Phosphoric acid does not contain P−H{P}-{H} bonds.
View answer and explanation
Correct option: A
Both (A) and (R) are true and (R) is the correct explanation of (A)
Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because phosphoric acid does not have P-H bonds, which are responsible for reducing properties in other phosphorus acids.
194 Medium
194 of 320
Understanding +4 | -1
Observe the following statements
I. Bleaching powder is used in the preparation of chloroform.
II. Bleaching powder decomposes in the presence of CoCl2\mathrm{CoC}{\mathrm{l}_\mathrm{2}} to liberate O2{\mathrm{O}_2}
III. Aqueous KHF2\mathrm{KH}{\mathrm{F}_\mathrm{2}} is used in the preparation of fluorine.
View answer and explanation
Correct option: D
Only I and II are correct
Option C (Only I and III are correct) is incorrect, so Option D (Only I and II are correct) is correct because aqueous KHF2 is NOT used in the preparation of fluorine.
195 Medium
195 of 320
Analyzing +4 | -1
Assertion (A): When H2S(g){H}_2 {S}({g}) is passed into aqueous ZnCl2{ZnCl}_2, solution, ZnS{ZnS} is precipitated.

Reason (R) : ZnS{ZnS} is insoluble in dil. HCl{HCl}.
View answer and explanation
Correct option: D
A\mathbf{A} and R\mathbf{R} both are false.
Option D (A\mathbf{A} and R\mathbf{R} both are false) is correct because ZnSZnS does not precipitate when H2SH_2S is passed into ZnCl2ZnCl_2 solution, and ZnSZnS is soluble in dilute HClHCl.
196 Medium
196 of 320
Remembering +4 | -1
Which of the silver salts are soluble in water?
View answer and explanation
Correct option: C
AgFAgF
Option C (AgFAgF) is correct because silver fluoride is soluble in water due to the high hydration enthalpy of the fluoride ion.
197 Medium
197 of 320
Remembering +4 | -1
Which of the following is used to provide smoke screens :
View answer and explanation
Correct option: A
Calcium phosphide
Option A (Calcium phosphide) is correct because calcium phosphide reacts with water to produce phosphine gas, which spontaneously ignites and creates smoke.
198 Medium
198 of 320
Analyzing +4 | -1
Assertion : Dinegative anion of oxygen (O2−)\left({O}^{2-}\right) is quite common but dinegative anion of sulphur (S2−)\left({S}^{2-}\right) is less common.

Reason : Covalency of oxygen is two.
View answer and explanation
Correct option: C
If the Assertion is correct but Reason is incorrect.
The correct answer is C (If the Assertion is correct but Reason is incorrect) because while the dinegative oxygen anion is common, the reason given (covalency of oxygen is two) is not the primary explanation. The stability of O2- is due to oxygen's high electron gain enthalpy.
199 Medium
199 of 320
Analyzing +4 | -1
The gaseous decomposition of ozone 2O3→3O22 \mathrm{O}_3 \rightarrow 3 \mathrm{O}_2 follows the mechanism

O3⇌keqO2+O (Fast), O+O3→k12O2 (Slow) \mathrm{O}_3 \stackrel{\mathrm{k}_{\mathrm{eq}}}{\rightleftharpoons} \mathrm{O}_2+\mathrm{O} \text { (Fast), } \mathrm{O}+\mathrm{O}_3 \xrightarrow{\mathrm{k}_1} 2 \mathrm{O}_2 \text { (Slow) }

Determine the order of reaction with respect to O2\mathrm{O}_2 :
View answer and explanation
Correct option: A
-1
Option A (-1) is correct because the rate law derived from the given mechanism shows that the reaction order with respect to O2 is -1.
200 Medium
200 of 320
Analyzing +4 | -1
A mixture of which pair of species react with water to produce a pure colourless gas that gives white fumes with HClHCl?
View answer and explanation
Correct option: C
Magnesium nitride and calcium nitride
Option C (Magnesium nitride and calcium nitride) is correct because both react with water to produce ammonia gas, which forms white fumes with HCl.
201 Medium
201 of 320
Analyzing +4 | -1
Assertion (A) : Cl2{Cl}_2 or Br2{Br}_2 changes KI{KI} into I2{I}_2 (violet in CCl4{CCl}_4 layer), Cl2{Cl}_2 changes KBr{KBr} into Br2{Br}_2 (orange-yellow in CCl4{CCl}_4 layer). Thus, Cl2{Cl}_2 would change mixture of KI{KI} and KBr{KBr} into violet layer in CCl4{CCl}_4.

Reason (R) : Oxidising power is in order I2<Br2<Cl2<F2{I}_2 < {Br}_2 < {Cl}_2 < {F}_2.
View answer and explanation
Correct option: A
Both A\mathbf{A} and R\mathbf{R} are true and R\mathbf{R} is the correct explanation of A\mathbf{A}.
Option A (Both A and R are true and R is the correct explanation of A) is correct because chlorine is a stronger oxidizing agent than bromine, which is stronger than iodine, thus chlorine will oxidize both iodide and bromide to their respective halogens.
202 Medium
202 of 320
Analyzing +4 | -1
Assertion (A): Conc. Sulphuric acid can be used to prepare HClHCl on reaction with NaCl.NaCl.

Reason (R): Conc. H2SO4\mathrm{H}_2 \mathrm{SO}_4 is a moderately strong reducing agent.
View answer and explanation
Correct option: C
(A) is true but (R) is false
Option C is correct because concentrated sulfuric acid can be used to prepare HCl, but it acts as an oxidizing agent, not a reducing agent.
203 Medium
203 of 320
Remembering +4 | -1
Which of the following pairs is obtained on heating ammonium dichromate?
View answer and explanation
Correct option: A
N2 and H2O{\mathrm{N}_\mathrm{2}}\,\mathrm{and}\,\mathrm{ }{\mathrm{H}_\mathrm{2}}\mathrm{O}
Option A (N2 and H2O{\mathrm{N}_{\mathrm{2}}}\,\mathrm{and}\,\mathrm{ }{\mathrm{H}_{\mathrm{2}}}\mathrm{O}) is correct because ammonium dichromate on heating decomposes to give nitrogen gas and water as products.
204 Medium
204 of 320
Understanding +4 | -1
On reaction with water,

I. fluorine oxidises water to oxygen.

II. chlorine and bromine react with water to form corresponding hydrohalic and hypohalous acids.

III. reaction of iodine with water is spontaneous.

Which of the above mentioned statements is/are correct?
View answer and explanation
Correct option: C
I and II
Option C (I and II) is correct because fluorine oxidizes water to oxygen, and chlorine and bromine react with water to form corresponding hydrohalic and hypohalous acids. Iodine's reaction with water is non-spontaneous.
205 Medium
205 of 320
Analyzing +4 | -1
Read the following statements regarding chemical reactivity of group 15 elements.

(i) Only compound of Bi{Bi} with +5 oxidation state is BiF5{BiF}_5.

(ii) Intermediate oxidation states for both nitrogen and phosphorus disproportionate in both acid and alkali.

(iii) Nitrogen due to absence of d-orbitals in its valence shell cannot form dπ{d} \pi-p π\pi bond as the heavier elements thus R3P=O{R}_3 {P}={O} exists but R3 N=O{R}_3 {~N}={O} does not exists.

(iv) BiH3{BiH}_3 is the strongest reducing agent amongst the hydrides of nitrogen family.

(v) P2O3{P}_2 {O}_3 is more acidic than P2O5{P}_2 {O}_5.

Which of the following is the correct code for the statements above?
View answer and explanation
Correct option: C
TFTTF
Option C (TFTTF) is correct because statements (i), (iii), and (iv) are true, while (ii) and (v) are false based on the chemical reactivity of group 15 elements.
206 Medium
206 of 320
Remembering +4 | -1
NH3  N{H_3}\; gas is dried over:
View answer and explanation
Correct option: C
Quick lime
Option C (Quick lime) is correct because ammonia gas can be dried over quick lime (CaO) as it doesn't react with it.
207 Medium
207 of 320
Understanding +4 | -1
The reaction in the Kipp’s apparatus stops on closing the outlet, because:
View answer and explanation
Correct option: D
The contact between sulphide and the acid is broken by the presence of gas collected in the free surface of the middle chamber
Option D (The contact between sulphide and the acid is broken by the presence of gas collected in the free surface of the middle chamber) is correct because the pressure of the gas produced builds up and pushes the acid away from the solid reactant, stopping the reaction.
208 Medium
208 of 320
Analyzing +4 | -1
Statement-I: P2O3{P}_2 {O}_3 is more basic than N2O3{N}_2 {O}_3

Statement-II: Metallic nature of the central atom increases down the group
View answer and explanation
Correct option: A
Statement I and II are correct
The correct answer is A because the basicity of oxides decreases down the group, and the metallic nature of the central atom increases down the group.
209 Medium
209 of 320
Remembering +4 | -1
The gas which is absorbed by ferrous sulphate solution giving blackish brown colour is:
View answer and explanation
Correct option: D
NO
Option D (NO) is correct because ferrous sulphate solution absorbs nitric oxide (NO) to form a blackish-brown colored complex.
210 Medium
210 of 320
Analyzing +4 | -1
Which of the following does not exist?
View answer and explanation
Correct option: B
[KrF3]−[SbF4]+{\left[ {Kr{F_3}} \right]^ - }{\left[ {Sb{F_4}} \right]^ + }
Option B is correct because KrF2 acts as a F- donor and forms complexes with F- acceptors, thus only cationic species or Kr will be present.
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